Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 15D / b

For two nonorthogonal input states $|\alpha ⟩,|\beta ⟩$, preservation of inner products by $G$ givesHence the two output-program states agree up to phase. Any two qubit states can be joined through a state nonorthogonal to both, so after absorbing phases
$|{\tilde{P}}_U⟩$ is independent of the input.

Inner-product preservation between programs for $U$ and $V$ givesfor all $|\alpha ⟩,|\beta ⟩$. If $⟨P_U|P_V⟩\ne 0$, this identity forces every matrix element of $U^{†}V$ to be the same scalar multiple of the corresponding identity matrix element. Hence $U^{†}V=e^{i\theta }I$, contrary to physical distinctness. Therefore

There are uncountably many physically distinct one-qubit unitaries. Part (ii) would require an orthogonal program state for every one, whereas a $K$-qubit register has dimension only $2^K$. This contradiction is the no-programming theorem, so no deterministic gate satisfying (PROG) exists.

Yes. Store $U$ in the two-qubit programApply the teleportation Bell measurement to the input and the first program qubit. On the ${\varphi }^{+}$ outcome, which has probability $1/4$ independently of $U$ and $|\alpha ⟩$, the remaining qubit is exactly $U|\alpha ⟩$. Accept that heralded outcome and declare failure otherwise. This gives a probabilistic universal programmable gate with constant success probability $p=1/4$.